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Two Polaroids are aligned so that the initially unpolarized light passing through them is a maximum. At what angle should one of

steposvetlana [31]

To solve this problem it is necessary to apply the law of Malus which describes the change in the Intensity of Light when it crosses a polarized surface.

Mathematically the expression is given as

$I = I_0 cos^2\theta$

Where,

$I_0$ = Initial Intensity

I = Final Intensity after pass through the polarizer

$\theta$ = Angle between the polarizer and the light

Since it is sought to reduce the intensity by half the relationship between the two intensities will be given as

$\frac{I}{I_0} = \frac{1}{2}$

Using the Malus Law we have,

$I = I_0 cos^2\theta$

$cos^2\theta = \frac{I}{I_0}$

$cos^2\theta = \frac{1}{2}$

$\theta = cos^{-1}(\frac{1}{2})^2$

$\theta = 75.52\°$

Angle with respect to maximum is $90-75.52 = 14.48\°$

8 0

3 years ago

A person is attracted towards the center of the earth by an 800 N gravitational force. The force with which the earth is attract

suter [353]

Answer:

800 N

Explanation:

By Newton's third law which states that for every action, there is an equal and opposite reaction.

So, as the earth attracts the person towards its center, the person attracts the earth towards itself with the same magnitude of force but in the opposite direction.

Since the person is attracted towards the center of the earth by an 800 N gravitational force, the the earth is attracted toward the person with an 800 N reaction force.

7 0

3 years ago

How much water will flow in 30 secs through 200 mm of capillary tube of 1.50 mm in diameter, if the pressure difference across t

Paladinen [302]

The water outflow in 30 secs through 200 mm of the capillary tube is mathematically given as

$Qo=1.6 \times 10^{2} \mathrm{~mL}$

<h3>What is the water outflow in 30 secs through 200 mm of the capillary tube?</h3>

$\begin{aligned}\Delta P &=6660 \mathrm{~m} / \mathrm{m}^{2} \\\mu &=8.01 \times 10^{-4} \text { Pas } \\t &=30 \mathrm{~s} \\L &=200 \mathrm{~mm}=200 \times 10^{-3} \mathrm{~m} \\D &=1.5 \mathrm{~mm}=1.5 \times 10^{-3} \mathrm{~m} \Rightarrow \gamma=\frac{1.5 \times 10^{-3}}{2} \mathrm{~m}\end{aligned}$